Find one value of $x$ that is a solution to the equation: $(x^2+3)^2+21=10x^2+30$ $x=$
Explanation: We could solve for $x$ by expanding $(x^2+3)^2$, combining terms that are alike, and using the quadratic formula or factoring to solve for $x$. However there is a more elegant way to approach this problem. Let's use structural features to rewrite the equation in a simpler form. Note that $10x^2+30=10({x^2+3})$. This means that we can rewrite the equation as: $({x^2+3})^2+21=10({x^2+3})$ If we let ${p}={x^2+3}$, we can see that this equation is in the form: ${p}^2+21=10{p}$ Let's solve this equation in terms of ${p}$ : $\begin{aligned}{p}^2+21&=10{p}\\\\ {p}^2-10{p}+21&=0\\\\ ({p}-3)({p}-7)&=0\\\\ {p}=3\ &\text{or} \ \ {p}=7 \end{aligned}$ Since ${p}={x^2+3}$, let's substitute this value back into our two solutions in order to solve for $x$ : ${x^2+3}=3\ \ \ \text{or} \ \ \ {x^2+3}=7$ When we solve ${x^2+3}=3$, we find that $x=0$. When we solve ${x^2+3}=7$, we find that $x=2$ or $x=-2$. In conclusion, the three solutions of the equation $(x^2+3)^2+21=10x^2+30$ are $x=0$, $x=-2$, and $x=2$. [Is there another way to solve for x?]